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8) A sample of agon has volume of 205 cm' when its temperature is 444.,0 "Cand its pressure is 712 mmHg At what temperature will the gas have volume of 23...

Question

8) A sample of agon has volume of 205 cm' when its temperature is 444.,0 "Cand its pressure is 712 mmHg At what temperature will the gas have volume of 235 cm' and pressure oF 850 mmHg?A fask: contains moles of carbon dioxide at pressure of 4tm Some carbon dioxide is removed from the flask and the new pressure is 2 atm, How many moles of gas are still in the flask? 10) student collects 00 x |0? cm' sample of hydrogen at 22,0 "€ and 91 kPa At what pressure will ithave t

8) A sample of agon has volume of 205 cm' when its temperature is 444.,0 "Cand its pressure is 712 mmHg At what temperature will the gas have volume of 235 cm' and pressure oF 850 mmHg? A fask: contains moles of carbon dioxide at pressure of 4tm Some carbon dioxide is removed from the flask and the new pressure is 2 atm, How many moles of gas are still in the flask? 10) student collects 00 x |0? cm' sample of hydrogen at 22,0 "€ and 91 kPa At what pressure will ithave temperature of 55.0"€ and volume of 4 60 10' cm "? H)A syringe contains 60 ml of air at temperature Of 30'C If the temperature is decreased (0 -308C , what is the new volunie? 12) volume of 2 [ aI # pressure Of ALITI , What will its volume be if the sample of gas occupies pressure I5 incrensed t0 WLn > volume of 88 ml 4t # pressure ol 40O (OrT. At what pressure will this sample 13) If a sample of gus has have volume 0f 200 mi? volume of 50 mL at & temperature of 273 K What will the volume be at a temperature 14) A ga5 occupies of 253 K? 15) sample of 388 ml sample helium has pressure of' 9 atm at temperature of' IS0K What will its volume be wien the pressure is 895 torr and temperature is 250*€ 16) Ni 3H: ZNH: How many liters of NH; are produced when 15 grarns of Nz react at standard temperature and pressure?



Answers

Answer the following questions:

(a) Is the pressure of the gas in the hot air balloon shown at the opening of this chapter greater than, less than, or equal to that of the atmosphere outside the balloon?
(b) Is the density of the gas in the hot air balloon shown at the opening of this chapter greater than, less than, or equal to that of the atmosphere outside the balloon?
(c) At a pressure of 1 atm and a temperature of 20 °C, dry air has a density of 1.2256 g/L. What is the (average) molar mass of dry air?
(d) The average temperature of the gas in a hot air balloon is $1.30 \times 10^{2} \circ F$ Calculate its density, assuming the molar mass equals that of dry air.
(e) The lifting capacity of a hot air balloon is equal to the difference in the mass of the cool air displaced by the balloon and the mass of the gas in the balloon. What is the difference in the mass of 1.00 L of the cool air in part (c) and the hot air in part (d)?
(f) An average balloon has a diameter of 60 feet and a volume of $1.1 \times 10^{5} \mathrm{ft}^{3}.$
What is the lifting power of such a balloon? If the weight of the balloon and its rigging is 500 pounds, what is its capacity for carrying passengers and cargo?
(g) A balloon carries 40.0 gallons of liquid propane(density 0.5005 $\mathrm{g} / \mathrm{L} )$ . What volume of $\mathrm{CO}_{2}$ and $\mathrm{H}_{2} \mathrm{O}$ gas is produced by the combustion of this propane?
(h) A balloon flight can last about 90 minutes. If all of the fuel is burned during this time, what is the approximate rate of heat loss (in kJ/min) from the hot air in the bag during the flight?

So continuing on with some of our walk around our physical chemistry. The fasting we're looking to solve here is for a temperature, so just drop some contextual information. The equation that we need T degrees Celsius says, has already been rearranged people to be over a minus. Log P minus c So we solved t t degrees c Goto 1750.286 Divide that by 8.10765 Tick with log 50 millimeters of mercury. Take away 235 Temperature is able to 38.1 degrees Celsius. A constant pressure and temperature The first drop of liquid forms of 38.1 degrees Celsius, so moving on to the next part. So the gas temperature is given it's 50 degrees Celsius. We've got a pressure and then we can calculate temperature. T is equal to 50 degrees Celsius. Add to 73.15 32 3.15 Calvin. Now we can calculate the number of models for gas mixtures with given temperature, so we're using the ideal gas equation P V equals and R T sofa, and that is not 0.744 moles. So in these miles of Catholics attempts and mile of water vapors in there, so most of water vapor equal to no point No. 744 So the remaining moles of gas our end too, and on to that is not 0.6696 moles. So the most of water vapor condensed will be not point. Not 5 to 08 And then we can use stands with water obtained on the models if the condensed water to calculate the volume occupied. So we have. Okay, liquid water is equal. Two months Bye bye density than what we get is B. Liquid water is equal to not 0.94 centimeters. Cute. So next what we're solving for we need to following equation, we have log Peace Star is equal to a minus B of T f C, and we're plugging in values on solving for temperature now, so temperature after Avery arranged is one 750 1750 point to a six that's divided by 8.10765 Take with log 15 millimeters of mercury. Take away 235 Gives the body of 17.5 degrees Celsius. So if the gas mixtures extraordinary rigid walled container and the low pressure where the front moves in, the barometric pressure drops, then what we'll find is the density of the gas mixture will not change. Since the volume on the number of miles of gas mixture is not changed by change of pressure, the athlete pressure of the mixture will not change by biometric pressure changes. The partial pressure of the gas will not change by the change enviroment trick pressure. The gauge pressure of the gas will be changed by the changing of biometric pressure since its function of biometric pressure, more fraction of the gas will not change as the more mass and moles are not changed. And the two point is a function off mole fraction because the mole fraction is not changed by barometric pressures or do point will also not change by thes same barometric pressures

So let's start with the first part. The question. What is the pressure of the vapor within the flask? Well, we know the pressure of the room is 752 millimeters of mercury, and this is gonna be equal to the pressure in the flask because the pressure on the outside of a container is gonna be directly proportional to the pressure off the inside of the container. So we're going to use the same value, but we have to convert it into atmospheres. Well, one, we know that one atmosphere is equal to 760 millimeters of mercury. So all we need to do is take our initial value. So 252 and then multiply it using dimensional analysis and cross out like units. And we get 0.9894 atmospheres. We don't round it because we're gonna be using this later. Okay, that's the first one done moving on to be What is the temperature of the vapor in Kelvin? And what's the volume of the flask? Well, for temperature, we already know that this flask has been boiling at 99.7 Celsius. So that's gonna be the temperature that it's at. So all we need to do is add 273 to this number to get degrees in Children. And we get 372 point certain Kelvin. And now for volume. We know that the flask is already 248.1 mL. So all we have to do is divide this number by 1000, and that gives us 0.248 leaders. Okay, Now we need the mass of the vapor. And for this one, we have to take the mask off the empty flask and then subtracted by the mass off the field flask. The last when it's filled with vapor. And so we know initially it is 55.844 g. Yeah, and then when it's filled, it is 56.11 Actually, this would be easier if you switch these around, because that way we don't have to convert from a negative number. Okay, so then we subtract these two numbers and we get 0.257 grams of Veber steps. Okay, so now we need the moles of vapor and this is going to be solved using the ideal gas law. And the ideal gas law is written as so PV equals and r t, where t is temperature. Are is the Ray Bourque's constant and in this case, it's going to be 0.8 to 1. Although this depends on what units you're using and lower case n that's moles these volume and please pressure. So all we need to do is just rearrange this equation to solve for moles. And all we need to do now is just plug in the numbers that we got earlier. So pressure pressure waas zero point 9894 So all right, that volume waas 0.248 are is 0.8 to 1 and the temperature was 372 0.7. So then we calculate all of these. Okay? Okay, okay. And we get about point 000 eight or 19 most Now I'm making this very long. Just so when we round it were sure of what number we are going to stick with now. We need the molar mass off one mole of vapor and as we know Moeller Mass is equal to the mass of the object minus and animals. So we know that the mass waas 0.257 and the moles is a very small number. So now we just calculate this, okay? And we get 320 0.5 g per mole and then go as far your answers.

Hello Today we'll be talking about Chapter 11 Question 66 which asks us to calculate Cem pressures of gases given other components of an equation. So here we have a gas that is eight leaders in volume. The temperature is 320 Kelvin and we have 1.3 moles of gas. And so if we recall our ideal gas equation, PV equals N R. T. Where are is just a constant coefficient for a just a constant that relates the units, we're told the volume, the number of moles, the constant and the temperature. And we were asked to solve for the pressure. We can accomplish this by simply dividing both sides by volume, and we get an R T divided by volume equals the pressure. And so with this, we can plug in the numbers that were given. We can write 1.3 moles times 0.8 to 06 Leader atmospheres Her mole Kelvin Times 320 Kelvin And then all of that is divided by eight liters. And that's going to equal or pressure. In terms of units, we can see that moles are going to cancel. Kelvin will cancel their on this top row of the fraction here leaders is also going to cancel. That will leave us with atmospheres is our final units. And so if we just plug all of these numbers indoor calculator, we can find the pressure is equal to four 0.27 atmospheres. And all we had to do was used the ideal gas equation to sulfur unknown variable. We can try this again with the same equation. PV equals n r t. And once again, we can see that we're given the volume temperature, constant and the number of moles and asked to find the pressure, the same rearrangement. And our tea over V affords this pressure. And so if we go to plug in numbers now, we can have actually do that down here. P The pressure is going to equal, uh, 1.5 moles time 0.8 to 06 leader atmospheres per mole Kelvin Times 340 Kelvin and all that divided by 10 liters. And so once again, all of our units will cancel, except for the pressure units, an indication that we're doing this correctly and we find that P is equal to four 0.19 atmospheres a T M. Oops. And so lastly, we have one more example to, ah, practice using the ideal gas equation. This is the ideal gas law. And so we have two gases in cylinders, let's say, and were asked to determine which of them has a larger pressure. And so for gas one, we're told that we have 1.3 moles of gas. We have the gas is at 300 kelvin, both gases or 300 telephone. So both gases or 300 Kelvin. And finally, the volume of the first gas is at 7.5 leaders. And of course, our never changes 0.8 to 06 leader atmospheres, her mole Kelson. And so if we just write that over here leader atmospheres for more Kelvin, we have the number of moles given to us for the second gas, which is 0.5 moles, and we have a volume of 10 leaders. And so we have the same equation. We're searching for P, and thus that gives us the equation that we've been using on the previous slides. P equals N. R. T over the and so if we want to solve for, ah, the first gas P one and move everything over and we can see that we can just plug in all of our numbers we get P one is going to equal 1.3 moles times our which I'm not gonna write out just for space purposes. And then times 300 Kelvin divided by 7.5 meters, soapy one, it is equal to 4.27 atmospheres. He too, on the other hand, so p two, it's going to be equal to 0.5 moles. Times are times 300 Kelvin. This is going to be divided by 10 liters and soapy to again just plugging it all indoor calculator equal toe one point to three atmospheres. And so we can now say that p two are that the second gas is at a lower pressure than the first gas. Just by calculating the pressure using the ideal gas law PV equals an arty or P equals and Artie over V and using the given amounts so lower pressure. And so hopefully this helps you become a little bit more comfortable using the ideal gas law to calculate pressures. Think

Okay, so I have 159 millimeters of mercury. Yeah, At 20°C. And I have 1 65 millimeters of mercury At 30°C. Mm. So we want to decide whether we have a liquid gas or if the situation there describing is simply impossible. So if my pressure is going to register 256 millimeters of mercury, well, that's impossible. Okay, I can't be it can't be greater than the vapor pressure. So you're measured pressure when it vapor liquid vapor rises, the pressure can't exceed the vapor pressure. So that's just impossible. Yeah. If we're It's gonna read less than our vapor pressure. So one, that means everything has vaporized. And we have only gas present, all the liquid has vaporized. Mhm. And then finally they're describing our vapor pressures. So we're giving us our vapor pressures. So we know that we have both a gas and a liquid present at that point.


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